Suppose that  (Subscript[E, 6])(z)  =  0  and  (Subscript[E, 7])(z)  =  0.   Then,  (Subscript[H, 7])(z) =  0  and the following computation for page 15 provides a check that :
     if (Subscript[E, 6])(z) =  (Subscript[E, 7])(z) =  0  and (Subscript[B, 1, 1])(z)  ≠  0, 
     then (1.77) and (1.75) yield (1.76).

The version number of Mathematica  used to evaluate this notebook:

$VersionNumber

Q2[z_] = y''[z]^2 + 2*c01[z]*y''[z]*y'[z] + 
          2*c02[z]*y''[z]*y[z] + c11[z]*y'[z] ^2 + 
          2  c12[z]*y'[z]*y[z] + c22[z]*y[z]^2;

B11[z_] = c11[z] - c01[z]^2;

B12[z_] = c12[z] - c01[z]*c02[z];

B22[z_] = c22[z] - c02[z]^2;

lam[z_] = 
        2*c01[z] - (D[B11[z], z] + 2*B12[z])/B11[z];

L = (y'''[
              z] + (c01[z] + lam[z]/2)*y''[
                z] + (D[ c01[z], z] + c02[z] + c01[z]*lam[z]/2 + 
                  B11[z])*y'[
                z] + (D[c02[z], z] + c02[z]*lam[z]/2 + B12[z])*
              y[z]);

expr1 = 
        D[Q2[z], z] + lam[z]*Q2[z] - 
            2*(y''[z] + c01[z]*y'[z] + c02[z]*y[z])*L  // 
          Together;

expr2 = Numerator[expr1]

expr3 = Coefficient[expr2, y[z]^2]

H7[z_] = 
      B11[z]*D[B22[z], z] - D[B11[z], z]*B22[z] - 2*B12[z]*B22[z] + 
          2*c01[z]*B11[z]*B22[z] - 2*c02[z]*B11[z]*B12[z] // 
        Expand

Expand[ expr3  -  H7[z] ]

expr4  =  Coefficient[ expr2,  y'[z]*y[z] ]

E6[z] = 
      B11[z]*B22[z] - 2*B12[z]^2 + c01[z]*B11[z]*B12[z] - c02[z]*B11[z]^2 + 
          B11[z]*D[B12[z], z] - D[B11[z], z]*B12[z] // Expand

Expand[expr4  -  2*E6[z]]

expr5  =  
      Expand[ expr2  -  H7[z]*y[z]^2  -  
          2*E6[z]*y[z]*y'[z] ]

This checks that the left and right members of (1.76) are equal.